Srinivas Physics

Tuesday, 12 March 2013

Physics 11th Thermal Properties of Matter

THERMAL PROPERTIES OF MATTER

1. Concept of Heat and Temperature

Familiar sensations of hotness and coldness are described with adjectives such as hot , warm, cold, cool etc.
When we touch an object, we use our temperature sence to ascribe to object a property called temperature.
Temperature of a body determines whether it feels hot or cold to the touch. The hotter you feel on touching the body higher is its temperature.
So we can say thet Temperature is relative measurement of hotness or coldness of a body.
An observation about hot and cold bodies in contact is that, when they both are in contact temperature of cold body increases and that of hot body decreases. This happens because energy is transferred from hot body to cold body when they are in contact and this is a nonmechanical process.
This energy which is transferred from one body to another without any mechanical work involved is known as HEAT.
Heat is a form of energy and heat transfer from one body to another takes place by virtue of temperature difference only also heat transfer takes place from body at higher temperature to body at lower temperature.
S.I. unit of heat is Joule(J) and that of temperature is Kelvin(K).

2. Measurement of temperature

Measurement of temperature can be obtained using a thermometer.
Construction of thermometers generally require a measurable property of a substance which monotonically changes with temperature.
Examples of some common type of thermometers
(1) Mercury in a glass thermometer.The height of mercury in the tube is taken as thermometric parameter.
(2) Constant Volume gas thermometer-Gas in bulb is maintaned at constant volume.The mean pressure of gas is taken as thermomtric parameter.
(3) Constant Pressure gas thermometer-Gas in bulb is maintaned at constant pressure.Volume of gas is taken as thermomtric parameter.
(4) Resistance therometer-Electric resistance of a metal wire increases monitonically with the temperature and may be used to define temperature scale. Such thermometers are resistance theromometers.
Thermometers are calibrated to asign a numerical value to any given temperature.
Defination of any standard scale needs two fixed refrence points and these points can be corelated to physical phenomenon reproducible at the same temperature.
Two such standard points are freezing and boiling points of water at same pressure.
Two such familiar scales used for measurement of temperature are Celsius and Fahrenhite scale.
Temperature in celsius is measured in degree.
Fahrenhite scale has a smaller degree then celsius scale and a diffrent zero of temperature.
Relation between Celsius and fahrenhite scale is
T_F=9/5 T_C + 32°.
where,
T_F - Fahrenhite Temperature.
T_C - Celsius Temperature.
Letters C & F are used to distinguish measurements on two scale thus
0° C= 32° F
this means that 0° C on celsius scale measures the same temperature as 32° F on the fahrenhite scale.
On fahrenhite scale melting point of ice and boiling point of water have values 32° F and 212° F and that on celsius scale are 0° C and 100° C.
If we now talk of Kelvin scale ,the melting point of ice and boiling point of water in the scale are 273.15 K and 373.15 K respectively.
Size of a degree in celsius and kelvin scale are same.
Relation between Celsius and kelvin scale is
T_C = T_K - 273.15 K
where,
T_K - Temperature in Kelvin
T_C - Temperature in celsius
Another modern fixed point of temperature is the triple point of water.
Three phases of water i.e, ice,water,water vapour co-exist at this value of temperature and pressure where
T_tr=273.16 K
P_tr= 0.46 cm of HG.
where,
T_tr & P_tr are triple point temperature and pressure.
Now if P_tr is pressure of an ideal gas thermometer at triple point temperature T_tr and if P is pressure at some other temperature T then corresponding temperature is
T= P( T_tr / P_tr)
provided T_tr & P_tr are low.
Electric resitance of metal wire increase monitonically with temperature and may be used to define the temperature scale
If R₀ & R₁₀₀ are resistance of metal wire at ice and steam point respectively then temperature t can be defined corresponding to resistance R_T as follows

T = (R_T-R₀)x100

R₁₀₀-R₀
A platinum wire is oftently used to construct a thermometer which is known as platinum resistance thermometer.
Gas thermometer can also define Celsius scale
If P₀ is pressure of gas at ice point and P₁₀₀₀ is pressure of gas at steam point then temperature T corresponding to a pressure P of gas is defined by

T = (P-P₀)x100

P₁₀₀-P₀

3. Absolute Temperature

In gas thermometer diffrent gases are used for measuring high temperatures and temperature reading are found to be independent of the nature of the gas used
Graphs between pressure and temperature for diffrent gases are plotted below
It is found that when graph were extrapolated for temperature below 0° C,pressure comes out to be zero at -273° C and this is the lowest temperature.
Lord kelvin suggested that instead of 0° C which is the melting point of ice,-273° C should be regarded as the zero of the temperature scale
Such a scale of temperature is absolute scale of temperature and -273.15° C as absolute zero of this new scale and is denoted as 0K and steam point in this scale correspond to 373.15 K

4. Ideal Gas Equation

Pressure of all the gases changes with the temperature in a similar fashion for low temperature.Also many properties of gases are common at low pressure
The pressure,volume and temperature in kelvin of such gases obey the equation
PV=nRT (1)
where n is amount of gas in number of moles and is given as ,
n = [Total no of molecules in given mass of gas]/[Avagadro Number(N_A)]
N_A=6.023 × 10²³
R is universal gas constant and its value is R=8.316 J/mol-K
Equation (1) is known as ideal gas equation and a gas obeying this equation is known as ideal gas.

5. Thermal Expansion

Most of the solid material expand when heated.
Increase in dimension of a body due to increase in its temperature is called thermal expansion.
For small change in temperature ΔT of a rod of length L, the fractional change in length ΔL/L is directly propertional to ΔT(Fig 3)

ΔL/L=α_LΔT (2)
or ,
ΔL=α_LLΔT (3)
Constant α_L characterizes the thermal expansion properties of a particulaqr material and it is known as coefficient of linear expansion.
For materials having no prefential direction,every linear dimension changes according to equation (3) and L could equally well represent the thickness of the rod,side lenght of the square sheet etc.
Normally metals expand more and have high value of α.
Again consider the intial surface area A of any surface and A' is the area of the solid when the temperature of the body changes by ΔT then increase in surface area is given by
ΔA=α_AAΔT (4)
where α_A is the coefficient of area expansion.
Similary we can define coefficient of volume expansion as fractional change in volume ΔV/V of a substance for a temperature change ΔT as

α_V = ΔV

VΔT
K^-1 is the unit of these coefficents of expansions.
These three coefficent are not strictly constant for a substance and there value is depends on temperaturerange in which they are being measured.
As an example, fig below shows that coefficient of volume expansion increase with temperrature and takes a constant value above 500K

Relation between volume and linear coefficient of expansion for solid material:

Consider a solid parallopide with dimension L₁,L₂ and L₃ then its volume is
V= L₁L₂L₃
When temperature increase by a amount ΔT then each linear dimension changes and then new volume is
     V+ΔV=L₁L₂L₃(1+α_LΔT)³
        =V(1+α_LΔT)³
        =V(1+3α_LΔT+3α_L²(ΔT)²+α_L³(ΔT)³)
if ΔT is small then higher order of ΔT can be neglected.Thus we find
     V+ΔV=V(1+3α_LΔT)
or,
     ΔV=3α_LV₀ΔT
Comparing this with equation (5) we find
α_V=3α_L

6. Thermal stress

If we fix the ends of a rod rigidly so that it can not expand or contract then with change in temperature, tensile or compressive stress known as thermal stresses will set up in the rod.
To compute the thermal stress,consider a rod of length L and crossectional area A with its both end rigidly fixed and temperature is then reduced by an amount ΔT.
The fractional change in rod if the rod is free to contract would be
ΔL/L=αΔT (6)
where ΔL and ΔT both are negative.
As we have fixed the ends of the rod, so it is not free to contract and this causes an increase in tension so as to produce an equal and opposite fractional change in length.
From the defination of Young Modulus

Y = F/A

ΔL/L

ΔL/L = F

AY

where ΔL is positive
Tensile forces is determined by the requirement that total fractional changes in length,Thermal expansion plus elastic strain must be zero.
     αΔT + F/AY=0
     F=-AYαΔT                         (7)
Thus tensile stress in rod is
     F/A=-YαΔT
Hence F/A is positive in case of decrease in temperature since change in temperature becomes negative.
In case where change in temperature tends to increase in temperature of the body then F and F/A becomes negative, corresponding development of to compressive force and strength.
Finaly stress which is is positive with decrease in temperature is Tensile stree and stress which is negative with increase in temperature is Compressive stress.

7. Specific Heat Capacity

If a system undergoes a change of temperature from T to T+ΔT during the transfer of ΔQ amount of heat then heat capacity c of the system is defined as the ratio of

c = ΔQ

ΔT
Thus Heat capacity per unit mass of a substance is its specific heat capacity.

c = ΔQ

mΔT

where,
m - mass of the substance
ΔQ - Heat absorbed or rejected by the substance
ΔT - Change in Temperature
Specific heat capacity depends on the nature of substance.
It is constant characterstics of the substance and is independent of the ammount of substance.
It also depends on the temperature of the substance
Its unit is J Kg^-1 K^-1
If the amount of substance is specified in terms of no of moles n instead of mass m then the heat capacity per mole of the substance is

C = ΔQ

nΔT

and is known as Molar Specific Heat capacity.
It is constant characterstics of the substance and independent of the ammount of substance
It depends on the nature of the substance ,temperature and amount of heat supplied
Its unit is J mol^-1 K^-1
In case of gases , when a gas is heated, ordinarly there is change in volume as well as pressure in addition to change in temperature
For simiplicity either volume or pressure can be kept constant.Thus gas have two specific heat capacities
1) Specific heat capacity at constant volume C_V
2) Specific heat capacity at constant pressure C_P

8. Calorimetery

Calorimetery means measurement of Heat.
Calorimeter is the device used to measure heat and it is cylinderical vessel made of copper and provided by a stirrer and a lid.
This vessel is kept in a wooden block to isolate it thermally from suroundings.A theromometer is used to measure the temperature of the content in the calorimeter.
When bodies at diffrent temperature are mixed in a calorimeter,they exchange heat with each other.
Bodies at higher temperature loose heat while bodies at low temperature gain heat.Contents of the calorimeter is continously stirred to keep temperature of contents uniform
Thus principle of calorimetery states that the total heat given by hot objects is equal to the total heat received by cold objects.

9.Change of phases:

There are three phases of matter i.e, Solid, Liquid and Gas.
Substance for example, H₂O exists in solid phase as Ice,in liquid phase as Water and in gas state as Steam.
Transtion from one phase to another phase are accompained by absorption or liberation of heat and usually by change in volume even at constant T.
Change of phase from solid to liquid is called melting , from liquid to solid is called fusion and from liquid to gas is called vaporisation
Once the temperature for phase change is reached( e,g melting or boiling temperature) no further temperature change occurs until all the substance has undergone phase change.

Melting Point:-Temperature at which solid and liquid phase are in thermal equilibrium with each other
Boiling point:-Temperature at which liquid and vapour phase are in thermal equilibrium with each other
Change of phase from solid state to Vapour state without passing through liquid state is called sublimation for e.g Dry Ice and Iodine sublimes.

10. Latent Heat:

The amount of heat per unit mass that must be transfered as heat when a sample completely undergoes phase change is called latent Heat of the substance for the process.
Thus when a sample of mass m completely undergoes phase change,the total energy transfered is
Q=Lm
where,
L- Latent heat and is characterstics of a substance
Its unit is JKg^-1.
Latent Heat for a solid liquid change is called Latent Heat of Fusion L_f.
Latent Heat for a liquid gas change is called Latent Heat of Vaporisation L_v.

Solved Examples

Question 1
A circular hole of diameter 2.00 cm is made in an aluminium plate at 0 ⁰ C .what will be the diameter at 100⁰ C?
Linear expansion for aluminium = 2.3 * 10^-3 / ⁰ C

Solution:
Diameter of circular hole in aluminium plate at 0⁰ C=2.0 cm
With increase in temperature from 0⁰ C to 100⁰ C diameter of ring increases

using
L=L₀(1+αΔT)
where L0=2.0 cm
α = 2.3 * 10^-3 / ⁰ C
ΔT=(100 -0)=100 ⁰ C
we can find diameter at 100⁰ C
L=2(1+2.3*10^-3*100)
=2.46 cm

Question 2
The pressure of the gas in constant volume gas thermometer are 80 cm,90cm and 100cm of mercury at the ice point,the steam point and in a heated wax bath resp.Find the temperature of the wax bath

Solution
Given that
Pressure at the ice point P_ice= 80 cm of Hg
Pressure at the steam point P_steam= 90 cm of Hg
Pressure at the wax bath P_wax= 100 cm of Hg
T=(P_wax-P_ice)X100/(P_steam-P_ice)
T = (100 - 80)X100/(90-80)
= 20X100/10
=200 ⁰C

Question 3
A rod of length L having coefficent of Linear expansion a is lying freely on the floor.it is heated so that temperature changes by b .Find the longitidunal strain developed in the rod
a. 0
b. ab
c. -ab
d. none of the above
Solution
There was no restriction for it expansion.So no tensile or compressive force developed.Longitudinal strains happens only when tensile or compressive force developed in the rod.So answer is a

Question 4
.if a is coefficent of Linear expansion,b coefficent of areal expansion,c coefficent of Volume expansion.Which of the following is true
a. b=2a
b. c=3a
c. b=3a
d. a=2b
Solution
Answer is c

Question 5
.which is of them is not used as the measurable properties in thermometer?
a.Resistance of platinum wire
b.Constant volume of gas
c.Contant pressure of gas
d.None of the above
Solution
Answer is d

Question 6
.when a solid metalic sphere is heated.the largest percentage increase occurs in its
a.Diameter
b. Surface area
c. Volume
d. density
Solution
Answer is c

Question 7
.the density of the liquid depends upon
a. Nature of the liquid
b. Temperature of the liquid
c. Volume of the liquid
d. Mass of the liquid
Solution
Answer is a and b
Question 8

A metallic sphere has a cavity of diameter D at its center.If the sphere is heated,the diameter of the . cavity will
a. Decrease
b. Increase
c. Remain unchanged
d. none of the above
Solution
Answer is b

Question 9
.A metallic circular disc having a circular hole at its center rotates about it axis passing through the center and perpendicular to it plane.when the disc is heated
a. Its speed will decrease
b. Diameter will increase
c. Moment of inertia will increase
d. its speed will increase
Solution
(a),(c)
Due to thermal expansion,the diameter of the disc as well of the hole will increase.therefore the moment of inertia will increase resulting in a increase in the angular speed.
Question 10
A resistance thermometer is such that resistance varies with temperature as
R_T=R₀(1+aT+bT⁵)
where T represent Temperature on Celsius scale And a,b,R₀ are constants.R₀ unit is ohm
Based on above data ,Find out the unit of a,b
Solution

As per dimension analysis Unit on both sides should be equal
Now since R & R₀ both unit are same
Quantity 1+aT+bT⁵ should be dimension less
so at should be dimension less
so a unit is C^-1
similarly bT⁵ should be dimensionless
so b unit is C^-5

Physics 11th Elasticity

ELASTICITY

1. Introduction

We know that any solid body has definite shape and size and more or less all solid bodies can be deformed by suitable application of forces.
Forces producing deformation in any solid body can bring about changes in length , volume or shape of the body and the body is said to be strained or deformed.
When deforming forces are removed the body tends to recover it's original condition.
This property of material body to regain it's original condition, on removal of deforming forces is called ELASTICITY.
When a solid body is deformed then it's constituents i.e., atoms or molecules gets displaced from their equilibrium position causing a change in interatomic or intermolecular distances.
Again when the deforming forces are removed interatimic forces drives atoms or molecules back to their original equilibrium position. This way body regains it's original shape and size.

DEFINITIONS
(a) Elastic forces:- The forces developed inside the body when deformed , tending to restore it's original shape are called elastic forces.
(b) Perfectly elastic bodies:- Bodies which which can recover their original condition completely on removal of deforming forces are called perfectly elastic bodies.
(c) Plastic bodies:- Bodies which does not show any tandency to recover their original condition on removal of deforming forces are called plastic bodies.
There are no perfectly elastic or plastic bodies and actual bodies lie between two extremes.
Nearest aproach to perfectly elastic body is a qquartz fibre and perfectly plastic body is putty.

2. Stress

We know that when deforming forces acts on a body, forces of internal reaction develops inside the body which tends to restore the body to it's original position.
These internal forces developed are equal in magnitude of deforming forces and acts in direction opposite to these externally applied deforming forces.
Stress is this restoring force applied per unit area set up inside the body and is measured by the magnitude of deforming force acting on unit area with in the elastic limits of the body.
Thus,

where F is the force applied and A is the area of crosssection of the body.
S.I. unit of stress is Nm^-2 or pascal (Pa). In C.G.S. system it's unit is dynes/cm².
Dimensional formula for stress is [ML^-1T^-2].
Stress are of two types

(a) Normal stress:- If elastic forces developed are perpandicular to the area of crossection of the body then the stress developed is known as normal stress.
The stress is always normal in case of change in length of wire or in case of change in volume of body shown below in figure
The normal stress are of two types, tensile and compressive stress, accordingly as there is a increase or decrease in length or volume of body on application of force.

(b) Tangential or shearing stress:- Tangential or shearing developes in a body when elastic restoring forces are parallel to the cross-sectional area of the body as shown below in figure 2.
Thus when deforming force acts tangentially over an area the body gets sheared through a certain angle.

3. Strain

When a body is under a system of forces or couples in equilibrium then a change is produced in the dimensions of the body.
This fractional change or deformation produced in the body is called strain.
Strain is a imensionless quantity.
Strain is of three types
(a) Longitudinal strain:- It is defined as the ratio of the change in length to the original length. If l is the original length and Δl is the change in length then,

(b) Volume strain:-It is defined as the ratio of change in volume to the original volume

(c) Shearing strain:- If the deforming forces produce change in shape of the body then the strain is called shear strain. Considering Figure 2. it can also be defined as the ratio of displacement x of corner b to the transverse dimension l. Thus

or,
Shear strain = tanθ
In practice since x is much smaller than l so, tanθ ≅ θ and the strain is simply the angle θ(measured in radians). Thus, shear strain is pure number without units as it is ratio of two lengths.

4. Hook's Law

Hook's law is the fundamental law of elasticity and is stated as " for small deformations stress is proportional to strain".
Thus,
stress ∝ strain
or,
stress/strain = constant
This constant is known as modulus of elasticity of a given material.
Hook's law is not valid for plastic materials.
Units and dimension of the modulus of elasticity are same as those of stress.

5. Elastic Modulus

Stress required to produce a given strain in a material body depends on the nature of material under stress.
We already know that ratio of stress to strain is known as elastic modulus of the material.
Larger is the elastic modulus of a given material, greater would be the stress needed to produce a given strain.
There are three different types of modulus of elasticity- Young's Modulus of elasticity, Bulk Modulus of elasticity and Modulus of Rigidity.
(a) Young's Modulus of Elasticity
Young's Modulus of elasticity is the ratio of longitudinal stress to longitudinal strain.
It is denoted by Y.
Young's Modulus of elasticity is given by

or
Let us now consider a wire of length l having area of cross-section equal to A. If the force F acting on the wire, stretches the wire by length Δ l then

and

From (1) and (2) we have Young's modulus of elasticity as
Young's modulus of elasticity has dimensions of force/Area i.e. of pressure.
Unit of Young's modulus is N/m².
If area of cross-section of a wire is given by A = πr² then Young's modulus is

again if A = π r² = 1cm² and Δ l = l = 1cm then
Y = F
Thus, Young's modulus can also be defined as the force required to double the length of a wire of unit length and unit area of cross-section.

(b) Bulk Modulus of Elasticity
The ratio of normal stress to volume strain within elastic limits is called Bulk Modulus of elasticity of a given material.
It is denoted by K.
Suppose a force F is applied normal to a surface of a body havin cross-sectional area equal to A.
If applied force bring about a change ΔV in the volume of the body and V is the original volume of the body then,

and

So, Bulk Modulus of elasticity would be,

Thus,
For gases and liquids the normal stress is caused by change in pressure i. e.,
normal stress = change in pressure ΔP.
Thus, bulk Modulus is

here negative sign indicates that the volume decreases if pressure increases and vice-versa.
For extremely small changes in pressure and volume, the Bulk Modulus is given by
Reciprocal of Bulk Modulus is called compressibility of substance. Thus,

(c) Modulus of Rigidity
When a body is sheared, the ratio of tangential stress to the shearing strain within elastic limits is called the Modulus of Rigidity.
If lower face of the rectangular block shown below in the figure, is fixed and tangential force is applied at the upper face of area A, then shape of rectangular block changes.

So,
shearing strain = θ ≅ tanθ
or,

Thus,

6. Poisson's Ratio

When two equal and opposite forces are applied to a body in a certain direction , the body extends along that direction and at the same time it cintracts along the perpandicular direction.
The fractional change in length of the body in the direction of the applied forces is longitudinal strain and fractional change in the perpandicular direction of the force applied is called lateral strain.
The ratio of lateral strain to the longitudinal strain is called poisson's ratio which is constant for material of that body.
So when a body is subjected to strain , say an elongation, it also suffers contraction in parpandicular direction.
Within elestic limits, lateral strain β is is proportional to longitudinal strain α.hence

σ = β

α

since,

longitudinal strain = α = Δl

l

and

lateral strain = β = ΔD

D
hence poisson's ratio is

σ = lΔD

DΔl

7. Stress-Strain Digram

In case of solids if we go on increasing stress continually then a point is reached at which strain increases more and more rapidly and Hook's law is no longer obeyed.
Thus, the stress at which linear relationship between stress and strain ceases to hold is referred as elastic limit of material for the stress applied.
If the elastic limit of material is exceeded it will fail to recover its original shape or size on removal of stress and would acquire a permanent set.
Any type of stress can be plotted against appropriate strain and the shape of resulting stress- strain digrams would have shapes, depending on the kind of material.
Simple stress- strain digram for a bar or wire is shown below in the figure.

(i). Portion OA is the straight line which clearly shows that stress produced is directly proportional to strain i.e., Hook's law is perfectly obeyed upto A and on removal of stress wire or bar will recover its original condition. Point A is called Proportionality limit

(ii). As soon as proportionality limit is crossed beyond point A, the strain increases more rapidly than stress and curve AB in graph shows that extension of wire in this limit is partly elastic and partly plastic and point B is the elastic limit of the material. Thus if we start decreasing load from point B the graph does not come to O via path BAO instead it traces straight line BG. So that there remains a residual strain. This is called permanent set.

(iii). If we continue to increases the stress beyond point B then for little or no increase in stress the strain increases rapidly upto point C.

(iv). Further increase of stress beyond point C produces a large increase in strain untill a point E is reached at which fracture takes place and from B to D material is said to undergo plastic flow which is irreversible.
Conclusion :

The wire exhibits elasticity from O to b and plasticity from b to d. If the distance between b and d is more, then the metal is ductile. If the distance between b and d is small, then metal is brittle.
The substances which break as soon as the stress is increased beyond elastic limit are called brittle substances eg: glass, cast iron, high carbon steel.
The substances which have a large plastic range are called ductile substances. Eg: copper, lead, gold, silver, iron, aluminium. Ductile materials can be drawn into wires. Malleable materials can be hammered into thin sheets. Eg: gold, silver, lead.

Solved Examples

Question 1. A block of gelatin is 60 mm by 60 mm by 20 mm when unstressed. A force of .245 N is applied tangentially to the upper surface causing a 5mm displacement relative to the lower surface.The block is placed such that 60X60 comes on the lower and upper surface. Find the shearing stress,shearing strain and shear modulus
a) (68.1 N/m²,.25,272.4 N/m²)
b) (68 N/m²,.25,272 N/m²)
c) (67 N/m²,.26,270.4 N/m²)
d) (68.5 N/m²,.27,272.4 N/m²)

Solution:

Shear stress=F/A=.245/36*10^-4 ==68.1 N/m²

Shear strain= tan?= d/h=5/20=.25

Shear modulus (S) =shear stress/shear strain=272.4 N/m²

Question 2.A steel wire of diameter 4mm has a breaking strenght of 4X10⁵N. The breaking strenght of similar steel wire of diameter 2 mm is a.1X10⁵N.
b.4X10⁵N.
c.16X10⁵N.
d. none of the these

Solution 2.

Breaking strenght is proportional to square of diameter,Since diameter becomes half,Breaking strenght reduced by 1/4/ Hence A is correct.

Question 3.What is the SI unit of modulus of elasticity of a substance?BR> a. Nm^-1BR> b. Nm^-2
c. Jm^-1BR> d. Unitless quantity

Solution 3
Answer is b

Question 4A thick uniform rubber rope of density 1.5 gcm^-3 and Young Modulus 5X1010⁶ Nm^-2 has a length 8 m. when hung from the celing of the room,the increase in length due to its own weight would be ? a. .86m
b. .2m
c. .1m
d. .096m
Solution 4 The weight of the rope can be assumed to act at its mid point. Now the extension x is proportional to the original lenghth L. if the weight of the rope acts at its mid point,the extension will be that produced by the half of the rope.So replacing L by L/2 in the expression for Young 's Modulus ,we have Y=FL/2Al or l=FL/2AY Since F=mg=ρV=ρAL Therefore l=gL²ρ/2Y Substituing the values,we get l=.096m

Physics 11th Gravitation

Gravitation

(1) Introduction :

In our dairy life we have noticed things falling freely downwords towards earth when thrown upwards or dispped from some hight.
Fact that all bodies irrespective of their masses are accelerated towards the earth mith a constant aceeleation was first recognized by galive (1564-1642)
The motion of celesh'al bodies such as moon. earth plametes etc. and attrachieve of moon towards earth and arth towards sun is an interasting subject of study since long time.
Now the question's what is the force that produces such acceleration is which earth attract all bodies towards the centre and what is the law governing this force.
Is this law is same for both earthly and weshal bodies.
Answer to this question was given by Newton as he declared that "laws of nature are same for earthly and weshal Bodies".
The force between any object falling freely towards earth and that between earth and moon are gowerned by the same laws.
Johnaase kepler (1571-1631) Studied the planetary motion in detail and formulated his three laws of planetary motion, which were available Universal law of grawitation.

(2) Kepler's Law :-
Kepler's law of planetary motion are :-

(i)Law of orbits :-
Each planet revolues around the sun in an elliptical orbit with sun at one of the foci of the ellipse asd shown in fig (a) below.
Fig (a) An ellipse traced by planet sevolary round the sun.

AO = a - Sewi major axis
BO = b - Sewi minor axis
P - hearest point between planet and sun k/as perihetion
A - farthest point between planet and sun apheiton.
(ii)Law of areas :-
The line joining planet and the sun sweeps equal area in equal intervals of time" [fig b]
This law follows from the observation that when planet is nearer to the sun its velocity increases and It appears to be slower when it is farther from the sun.
(iii)Law of periods :-
The squre of time period of any planet about the sun is propotional to the cube of the semi-major axis."

If T is the time period of semi major axis a of elliptical orbit then.
T² x a³ (1)
If T₁ and T₂ are time periods of any two planets and a₁ and a₂ being their semi major axis resp. then
T₁² x a₁³ = a₁³
T₂² x a₂³ a₂³ (2)
This question (2) can be used to find the time period of a planet, when the time period of the other
planet and the semi-major axis of orbits of two planets.

(3) Universal law of gravition :-

Everybody in the universe attracts every other body with a force which is directly proportional to the
product of their masses and invessly proportional to the square of distance between them.
Mathematically Newton's gravitation law is if F is the force acting between two bodies of masses M₁ and M₂ and the distance between them is R then majnitade of force is given as


F = Gm₁m₂

r²

In vector notation

F = Gm₁m₂(-r^)

r²

F = -Gm₁m₂r^

r²

     where G- universal gravitational constant
     r^ - unit vector from m₁ to m₂ and r^ = r₁^ - r₂^
3-     Gravitational force is attrachive constant is
     SI -           G= 6.67 x 10^-11 nm² kg^-2
     CGS -           G= 6.67 x 10^-8 dyn cm² g^-2
     Deminsional formula of C1 is [m^-1L³T^-2]

(4) Acceleration due to gravity of earth :-

Earth attracts every object lying an its surface towards its centre with a force known as gravitational towards its centre with a force known as gravitational pull or gravity.
Whenever force acts on any body it produces acceleration and in case of gravitation this acceleration produced under effect of gravity is known as accelesation due to gravity (g)
Value of accelesation due to gravity is independent of mass of the body and its value near surface of earth is 9.8 ms^-2
Expression for acceleration due to gravity
Consider mass of earth to be as M_E and its redius be R_E Suppose a body of mass M (much smaller then that fo earth) is kept at the earth surface. Force eseerted by earth on the body of mass m is


F = -GMM_E

R_E²

      The force for the body due to earth produces acceleration due to      gravity
(g) in the motion of the body. From Newton's Second law of motion
     f=mg
     from (4) and (s)

g = -GM_E

R_E²

     which is acceleration due to gravity at earth's surface.

(5) Acceleration due to gravity below and above the earth surface :-

(i) Above earth's surface

An object of mass m is placed at hight h above the earth's active in this object is

F = -GMM_E

(R_E+h)²

From this it can be concluded that value of g decreases as distance above surface of earth increases now,

g = -GM_E

R_E(1+h/R_E)²

g = g_o

(1+h/R_E)²

Where

g_o = GM_E

R_²
eqn (7) tells us that for small hight h above surface of earth. value of g decreases by factor (1-2h/R_E)
- for h<<R
g=g_o(1+h/R_E)^-2
g=g_o(1-2h/R_E) Expanding by Binomial theriom

(ii) Below the earth's surface

If one goes inside the earth surface the value of g again decreases

P = density of material of earth them
m=(4/3)(R_E)³P
From this acceleration due to gravity at earth’s Surface is

g = G(4/3)R_E³P

R_E²

g=(4/3)GR_EP                              (8)
g-acceleration due to gravity at depth D below earth's surface
-Body at depth d will experience force only due to portion of reduce (R_E-1d) of earth's
-outer spherical shell of thickness d will not exeperience any force
-M is mas of the portion of earth with radius (R_E-d) then

g = -GM

R_E²

M=(4/3)(R_E-d)³ P

g = -G4/3(R_E-d)³ P

(R_E²-d)²

g= (4/3)G (R_E-d)P                    (9)
Dividing epn (9) by (8)
g/g_o= (1-d/R_E)
or g=go (1-d/R_E)
nbsp;                       (10)
from epn (10) is cleaer that acceeleration due to gravity also decreases with depth.

Physics 11th Simple Harmonic Motion

Simple Harmonic Motion

1. Periodic motion

If a particle moves such that it repeats its path regularly after equal intervals of time , it's motion is said to be periodic.
The interval of time required to complete one cycle of motion is called time period of motion.
If a body in periodic motion moves back and forth over the same path then the motion is said to be viberatory or oscillatory.
Examples of such motion are to and fro motion of pendulum , viberations of a tuning fork , mass attached to a spring and many more.
Every oscillatory motion is periodic but every periodic motion is not oscillatory for example motion of earth around the sun is periodic but not oscillatory.
Simple Harmonic Motion (or SHM) is the simplest form of oscillatory motion.
SHM arises when force on oscillating body is directly proportional to the displacement from it's equilibrium position and at any point of motion , this force is directed towards the equilibrium position.

2. Simple Harmonic Motion (or SHM)

SHM is a particular type of motion very common in nature.
In SHM force acting on the particle is always directed towards a fixed point known as equilibrium position and the magnitude of force is directly proportional to the displacement of particle from the equilibrium position and is given by
F= -kx
where k is the force constant and negative sign shows that fforce opposes increase in x.
This force os known as restoring force which takes the particle back towards the equilibrium position , and opposes increase in displacement.
S.I. unit of force constant k is N/m and magnitude of k depends on elastic properties of system under consideration.
For understanding the nature of SHM consider a block of mass m whose one end is attached to a spring and another end is held stationary and this block is placed on a smooth horizontal surface shown below in the fig.
Motion of the body can be described with coordinate x taking x=0 i.e. origin as the equilibrium positionwhere the spring is neither stretched or compressed.
We now take the block from it's equilibrium position to a point P by stretching the spring by a distance OP=A and will then release it.
After we release the block at point P, the restoring force acts on the block towards equilibrium position O and the block is then accelerated from point P towards point O as shown below in the fig.
Now at equilibrium position this restoring force would become zero but the velocity of block increases as it reaches from point P to O.
When the block reaches point O it's velocity would be maximum and it then starts to move towards left of equilibrium position O.
Now this time while going to the left of equilibrium position spring is compressed and the block moves to the point Q where it's velocity becomes zero.
The compressed spring now pushes the block towards the right of equilibrium position where it's velocity increases upto point O and decreases to zero when it reaches point P.
This way the block oscillates to and fro on the frictionless surface between points P and Q.
If the distance travelled on both sides of equilibrium position are equal i.e. , OP=OQ then the maximum displacement on either sides of equilibrium are called the Amplitude of oscillations.

3. Equation of SHM

Consider any particle executing SHM with origin as it's equilibrium position under the influence of restoring force F=
kx , where k is the force constant and x is the displacement of particle from the equilibrium position.
Now since F= -kx is the restoring force and from Newton's law of motion force is give as F=ma , where m is the mass of the particle moving with acceleration a. Thus acceleration of the particle is
          a=F/m
           =-kx/m
but we know that acceleration a=dv/dt=d²x/dt²
⇒           d²x/dt²=-kx/m          (1)
This equation 1 is the equation of motion of SHM.
If we choose a constant φ=√(k/m) then equation 1 would become
d²x/dt²=-φ²x (2)
This equation is a differential equation which says that displacement x must be a funcyion of time such that when it's second derivative is calculated the result must be negative constant multiplied by the original function.
Sine and cosine functions are the functions satisfying above requirement and are listed as follows
          x=A sinωt                         (3a)
          x=A cosωt                         (3b)
          x=A cos(ωt+φ)                    (3c)
each one of equation 3a, 3b and 3c can be submitted on the left hand side of equation 2 and can then be solved for varification.
Convinently we choose equation 3c i.e., cosine form for representing displacement of particle at any time t from equilibrium position. Thus,
x=A cos(ωt+φ) (4)
and A , φ and φ are all constants.
Fig below shows the displacement vs. time graph for phase φ=0.

4. Characterstics of SHM
Here in this section we will learn about physical meaning of quantities like A, T, ω and φ.

(a) Amplitude

Quantity A is known as amplitude of motion. it is a positive quantity and it's value depends on how oscillations were started.
Amplitude is the magnitude of maximum value of displacement on either side from the equilibrium position.
Since maximum and minimum values of any sine and cosine function are +1 and -1 , the maximum and minimum values of x in equation 4 are +A and -A respectively.
Finally A is called the amplitude of SHM.

(b) Time period

Time interval during which the oscillation repeats itself is known as time period of oscillations and is denoted by T.
Since a particle in SHM repeats it's motion in a regular interval T known as time period of oscillation so displacement x of particle should have same value at time t and t+T. Thus,
          cos(ωt+φ)=cos(ω(t+T)+φ)
cosine function cos(ωt+φ) will repeat it's value if angle (ωt+φ) is increased by 2π or any of it's multiple. As T is the pime period
          (ω(t+T)+φ)=(ωt+φ)+2π
or,          T=2π/` = 2π√(m/k)               (5)
Equation 5 gives the time period of oscillations.
Now the frequency of SHM is defined as the numberof complete oscillations per unit time i.e., frequency is reciprocal of time period.
f=1/T = 1/2π(√(k/m)) (6)
Thus, ω=2`/T = 2`f (7)
This quantity ω is called the angular frequency of SHM.
S.I. unit of T is s (seconds)
f is Hz (hertz)
ω is rad s^-1 (radian per second)

(c) Phase

Quantity (ωt+φ) in equation (4) is known as phase of the motion and the constant φ is known as initial phase i.e., phase at time t=0, or phase constant.
Value of phase constant depends on displacement and velocity of particle at time t=0.
The knowledge of phase constant enables us to know how far the particle is from equilibrium at time t=0. For example,
If φ=0 then from equation 4
          x=A cosωt
that is displacement of oscillating particleis maximum , equal to A at t=0 when the motion was started. Again if φ=`/2 then from equation 4
          x=A cos(ωt+`/2)
           =Asinωt
which means that displacement is zero at t=0.
Variation of displacement of particle executing SHM is shown below in the fig.

5.Velocity of SHM

We know that velocity of a particle is given by
v=dx/dt
In SHM displacement of particle is given by
x=A cos(ωt+φ)
now differentiating it with respect to t
v=dx/dt= Aω(-sin(ωt+φ)) (8)
Here in equation 8 quantity Aω is known as velocity amplitude and velocity of oscillating particle varies between the limits ±ω.
From trignometry we know that
     cos²θ + sin²θ=1
⇒
     A² sin²(ωt+φ)= A²- A²cos²(ωt+φ)
Or
     sin(ωt+φ)=[1-x²/A²]            (9)
putting this in equation 8 we get,
From this equation 10 we notice that when the displacement is maximum i.e. ±A the velocity v=0, because now the oscillator has to return to change it’s direction.
Figure below shows the variation of velocity with time in SHM with initial phase φ=0.

6. Acceleration of SHM

Again we know that acceleration of a particle is given by
a=dv/dt
where v is the velocity of particle executing motion.
In SHM velocity of particle is give by,
v= -ωsin(ωt+φ)
differentiating this we get,

or,
a=-ω²Acos(ωt+φ) (11)
Equation 11 gives acceleration of particle executing simple harmonic motion and quantity ω² is called acceleration amplitude and the acceleration of oscillating particle varies betwen the limits ±ω²A.
Putting equation 4 in 11 we get
a=-ω²x (12)
which shows that acceleration is proportional to the displacement but in opposite direction.
Thus from above equation we can see that when x is maximum (+A or -A), the acceleration is also maximum(-ω²A or +ω²A)but is directed in direction opposite to that of displacement.
Figure below shows the variation of acceleration of particle in SHM with time having initial phase φ=0.

7. Total energy in SHM

When a system at rest is displaced from its equilibrium position by doing work on it, it gains potential energy and when it is released, it begins to move with a velovity and acquires kinetic energy.
If m is the mass of system executing SHM then kinetic energy of system at any instant of time is
K=(1/2)mv² (13)
putting equation 8 in 13 we get,
From equation (14) we see that Kinetic Energy of system varies periodically i.e., it is maximum (= (1/2)mω²A²) at the maximum value of velocity ( ±ωA) and at this time displacement is zero.
When displacement is maximum (±A), velocity of SHM is zero and hence kinetic energy is also zero and at these extreme points where kinetic energy K=0, all the energy is potential.
At intermediate positions of lying between 0 and ±A, the energy is partly kinetic and partly potential.
To calculate potential energy at instant of time consider that x is the displacement of the system from its equilibrium at any time t.
We know that potential energy of a system is given by the amount of work required to move system from position 0 to x under the action of applied force.
Here force applied on the system must be just enough to oppose the restoring force -kx i.e., it should be equal to kx.
Now work required to give infinitesimal displacement is dx=kx dx.
Thus, total work required to displace the system from 0 to x is

thus,

where, from equation 5 ω=√(k/m) and displacement x=A cos(ωt+φ).
-From equation 14 and 15 we can calculate total energy of SHM which is given by,
Thus total energy of the oscillator remains constant as displacement is regained after every half cycle.
If no energy is dissipated then all the potential energy becomes kinetic and vice versa.
Figure below shows the variation of kinetic energy and potential energy of harmonic oscillator with time where phase φ is set to zero for simplicity.

8. Some simple systems executing SHM

(A) Motion of a body suspended from a spring

Figure (6a) below shows a spring of negligible mass, spring constant k and length l suspended from a rigid support.
When a body of mass m is attached to this spring as shown in figure 6(b), the spring elongates and it would then rest in equilibrium position such that upward force F_up exerted by spring is equal to the weight mg of the boby.
If the spring is extended by an amount Δl ater attachment of block of mass m then in its equilibrium position upward force equals
      F_up=kΔl
also in this equilibrium position
      F_up=mg
or,      kΔl=mg
Again the body is displaced in upwards direction such that it is at a distance x above equilibrium position as shown in figure 6(c).
Now extansion of spring would be (Δl-x), thus upward force now exerted on the body is
F_up=k(Δl-x)
Weight of the body now tends to pull the spring downwards with a force equal to its weight. Thus resultant force on the body is
     F=k(Δl-x)-mg
      =mg-kx-mg
or,
     F=-kx                    (17)
From equation 17 we see that resultant force on the body is proportional to the displacement of the body from its equilibrium position.
If such a body is set into vertical oscillations it oscillates with an angular frequency
ω=√(k/m) (18)

(B) Simple pendulum

Simple pendulum consists of a point mass suspended by inextensible weightless string in a uniform gravitational field.
Simple pendulum can be set into oscillatory motion by pulling it to one side of equilibrium position and then releasing it.
In case of simple pendulum path ot the bob is an arc of a circle of radius l, where l is the length of the string.
We know that for SHM F=-kx and here x is the distance measured along the arc as shown in the figure below.
When bob of the simple pendulum is displaced from its equilibrium position O and is then released it begins to oscillate.
Suppose it is at P at any instant of time during oscillations and θ be the angle subtended by the string with the vertical.
mg is the force acting on the bob at point P in vertically downward direction.
Its component mgcosθ is balanced by the tension in the string and its tangential component mgsinθ directs in the direction opposite to increasing θ .
Thus restoring force is given by
F=-mgsinθ (19)
The restoring force is proportional to sinθ not to The restoring force is proportional to sinθ, so equation 19 does not represent SHM.
If the angle θ is small such that sinθ very narly equals θ then above equation 19 becomes
     F=-mgθ
since x=lθ then,
     F=-(mgx)/l
where x is the displacement OP along the arc. Thus,
     F=-(mg/l)x               (20)
From above equation 20 we see that restoring forcr is proportional to coordinate for small displacement x , and the constant (mg/l) is the force constant k.
Time period of a simple pendulum for small amplitudes is
Corresponding frequency relations are

and angular frequency
ω=√(g/l) (23)
Notice that the period of oscillations is independent of the mass m of the pendulumand for small oscillations pperiod of pendulum for given value of g is entirely determined by its length.

(c) The compound pendulum

Compound pendulum is a rigid body of any shape, capable of oscillating about a horizontal axis passing through it.
Figure below shows vertical section of rigid body capable of oscillating about the point A.
Distance l between point A and the centre of gravity G is called length of the pendulum.
When this compound pendulum is given a small angulr displacement θ and is then released it begins to oscillate about point A.
At angular displacement θ its center of gravity now takes new position at G'.
Weight of the body and its reaction at the support constitute a reactive couple or torque given by
τ=-mg G'B
=-mglsinθ (24)
Equation 24 gives restoring couple which tends to bring displaced body to its original position.
If α is the angular acceleration produced in this body by the couple and I is the moment of inertia of body about horizontal axis through A then the couple is
Iα=-mglsinθ
if θ is very small then we can replace sinθ≅θ, so that
α=-(mgl/I)θ (25)
From above equation (25) we se that pendulum is executing Simple Harmonic Motion with time period

9. Damped Oscillations

Fractional force, acting on a body opposite to the direction of its motion , is called damping force.
Damping force reduces the velocity and the Kinetic Energy of the moving body.
Damping or dissipative forces generally arises due to the viscosity or friction in the medium and are non conservative in nature.
When velocities of body are not high, damping force is found to be proportional to velocity v of the particle i.e.,
F_d=-γv (27)
where, γ is the damping constant.
If we take damping into consideration for an oscillator then oscillator experiences
(i) Restoring Force :- F=-kx
(ii) Damping Force :- F_d=-γv
where, x is th edisplacement of oscillating system and v is the velocity of this dispalcement.
Thus equation of motion of damped harmonic oscillator is

where,r=(γ/2m) and ω²=k/m
Solution of above equation is of the form
x=Ae^-rtcos(ω't+φ) (29)
where,
ω'=√(ω²-r²) (30)
is the angular frequency of the damped oscillator.
In equation 29 , x is a function of time but it is not a periodic function and because of the damping factor e^-rt this function decreases continously with time.

10. Driven or Forced Harmonic oscillator

If an extra periodic force is applied on a damped harmonic oscillator, then the oscillating system is called driven or forced harmonic oscillator, and its oscillations are called forced oscillations.
Such external periodic force can be represented by
F(t)=F₀cosω_ft (31)
where, F₀ is the amplitude of the periodic force and ω_f is th e frequency of external force causing oscillations.
Differential equation of motion under forced oscillations is
In this case particle will neither oscillate with its free undamped frequency nor with damped angular frequency rather it would be forced to oscillate with angular frequency ω_f of applied force.
When damped oscillator is is set in forced motion, the initial motion is combination of damped oscillation and forced oscillations .
After certain amount of time the amplitude of damped oscillations die out or becomec so small that they can be ignoredand only forced oscillationd remainand the motion is thus said to reached steady state.
Solution of equation 32 is
x=Acos(ω_ft+φ) (33)
where A is the amplituse of oscollation of forced oscillator and φ is the initial phase.
In case of forced oscillations both amplitude A and initial phase φ are fixed quantities depending on frequency ω_f of applied force.
Calculations show sthat amplitude

and initial phase
tanφ=-v₀/(ω_fx₀)
where, x₀ is displacement of particle at time t=0, the moment driven force is applied and v₀ is the velocity of the particle at time t=0.
When ω_f is very close to ω, then m(ω²-(ω_f)² would be much less than ω_fγ, for any reasonable value of γ, then equation 34 becomes
A=F₀/γω_f (35)
Thus the maximum possible amplitude for a given driven frequency is governed by the driving frequency and the damping ,and is never infinity.
This phenomenon of increase in amplitude when the driving force is close to natural frequency of oscillator is called RESONANCE.
Thus resonance occurs when frequency of applied force becomes equal to natural frequency of the oscillator without damping.

T =	(R_T-R₀)x100
	R₁₀₀-R₀

T =	(P-P₀)x100
	P₁₀₀-P₀

α_V =	ΔV
	VΔT

Y =	F/A
	ΔL/L